Question

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $$\left\\{a_{n}\right\\}=\left\\{(-1)^{n} \frac{n}{n+1}\right\\}$$

Short answer

The sequence diverges because it oscillates indefinitely.

Step-by-step solution

Understand the Sequence

The sequence given is \( \{a_n\} = \left\{(-1)^n \frac{n}{n+1}\right\} \). This sequence oscillates between positive and negative values as \((-1)^n\) alternates the sign based on whether \(n\) is even or odd.

Analyze the Oscillating Factor

Observe the factor \((-1)^n\), which changes the sign of each term, indicating that the sequence does not remain solely positive or negative.

Expression of the Term

The term at position \(n\) is \((-1)^n \frac{n}{n+1}\). As \(n\) becomes very large, analyze the behavior of \( \frac{n}{n+1}\).

Limit of \( \frac{n}{n+1} \) as \( n \to \infty \)

Calculate: \[ \lim_{{n \to \infty}} \frac{n}{n+1} = \lim_{{n \to \infty}} \left(1 - \frac{1}{n+1}\right) = 1 \] because \( \frac{1}{n+1} \to 0 \) as \(n\) approaches infinity.

Behavior of the Sequence

Since \( \frac{n}{n+1} \to 1 \), the sequence becomes \((-1)^n \times 1 = (-1)^n\) as \(n\) becomes very large. This implies that the sequence oscillates indefinitely between 1 and -1.

Conclusion on Convergence

The sequence does not converge to a single real number as \((-1)^n\) causes it to oscillate indefinitely. Therefore, the sequence diverges.

Key concepts

Oscillating Sequences

When we talk about oscillating sequences, we are referring to sequences whose terms do not settle down to a specific number but instead move back and forth. This behavior can be observed in the given sequence \(\{-1, 1, -1, 1, \ldots\}\), generated by the factor \((-1)^n\) which alternates the sign of each term based on the parity of \(n\).
This means the sequence will keep switching between positive and negative.
Let's break it down with an example:

• For \(n=1\), the term is \((-1)^1 = -1\)
• For \(n=2\), the term is \((-1)^2 = 1\)

This alternation continues indefinitely, making the sequence never settle at a single value. Instead, it keeps oscillating between two values, which is why it's called an oscillating sequence.

Limits in Calculus

The concept of limits is fundamental in calculus, helping us understand the behavior of sequences and functions as they get infinitely large or small.
In the context of sequences, we are interested in what value the terms of a sequence approach as \(n\) becomes very large. Let's consider the part \(\frac{n}{n+1}\) from our sequence:
When \(n\) increases towards infinity, the expression \(\frac{n}{n+1}\) evaluates to:
\[ \lim_{{n \to \infty}} \frac{n}{n+1} = \lim_{{n \to \infty}}\left(1 - \frac{1}{n+1}\right) = 1 \]
This tells us that as \(n\) gets very large, \(\frac{n}{n+1}\) gets closer and closer to 1. However, since we're multiplying by \((-1)^n\), the sequence's terms still oscillate, thus not converging to a single value.

Divergence of Sequences

A sequence is said to diverge when it fails to settle at a single real number or doesn't approach any particular limit.
In the case of the sequence \(\left\{(-1)^n \frac{n}{n+1}\right\}\), the terms try to approach 1 through \(\frac{n}{n+1}\). But, the alternating factor \((-1)^n\) causes these terms to oscillate endlessly between -1 and 1.
This indefinite bouncing between numbers ensures that the sequence never converges.
So, why is divergence important? It gives us insight into the behavior and nature of sequences that do not stabilize, which is crucial in identifying whether certain mathematical models or functions will behave predictably as their input grows without bound.
Divergent sequences like this one can often be spotted when terms neither come closer together nor get progressively smaller, instead, they continue to wander without pinning down to one spot.